# Let $|a|=m$ and $|b|=n$ and $gcd(m,n)=1$. Show that $\left \langle a \right \rangle \cap \left \langle b \right \rangle = \left \{ e \right \}$.

Can someone please verify whether my proof is correct?

Let $a,b\in G$ with $|a|=m$ and $|b|=n$ and $gcd(m,n)=1$. Show that $\left \langle a \right \rangle \cap \left \langle b \right \rangle = \left \{ e \right \}$.

Proof:

By the fundamental theorem of cyclic groups, if $|a|=m$ and $|b|=n$, then the order of $\left \langle a \right \rangle \cap \left \langle b \right \rangle$ is a divisor of $m$ and $n$. Then $|\left \langle a \right \rangle \cap \left \langle b \right \rangle|=gcd(m,n)=1$. Additionally, $\left \langle a \right \rangle \cap \left \langle b \right \rangle$ has exactly one subgroup of order $1$, which is the trivial subgroup $\left \{ e \right \}$. However, since there can only be exactly one such subgroup of that order, it must be that $\left \langle a \right \rangle \cap \left \langle b \right \rangle = \left \{ e \right \}$. $\square$

• I would say you could even end it after saying the order of the intersection must be one, because there is only one group of order one. – Dave Feb 5 '18 at 4:55

$\langle a\rangle\cap\langle b\rangle$ is a common subgroup of $\langle a\rangle$ and $\langle b\rangle$. According to the Lagrange Theorem, $|\langle a\rangle\cap\langle b\rangle |$ is a common divisor of both $|a|$ and $|b|$, namely $m$ and $n$. However, we have known that gcd$(m,n)=1$. Therefore, $|\langle a\rangle\cap\langle b\rangle |=1$.